Difference between revisions of "2018 AMC 10B Problems/Problem 17"
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Now notice that since <math>CD=8-2x</math> we have <math>QC=DR=x-1</math>. | Now notice that since <math>CD=8-2x</math> we have <math>QC=DR=x-1</math>. | ||
− | Thus by the Pythagorean Theorem we have <math>x^2+(x-1)^2=(8-2x)^2</math> which becomes <math>2x^2-30x+63=0\implies x=\dfrac{15-3\sqrt{11}}{2}</math> | + | Thus by the Pythagorean Theorem we have <math>x^2+(x-1)^2=(8-2x)^2</math> which becomes <math>2x^2-30x+63=0\implies x=\dfrac{15-3\sqrt{11}}{2}</math> |
Our answer is <math>8-(15-3\sqrt{11})=3\sqrt{11}-7\implies \boxed{\text{(B)}~7}</math>. (Mudkipswims42) | Our answer is <math>8-(15-3\sqrt{11})=3\sqrt{11}-7\implies \boxed{\text{(B)}~7}</math>. (Mudkipswims42) | ||
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==Solution 4== | ==Solution 4== | ||
− | Let <math>AB</math>, or the side of octagon, be <math>x</math>. Then, <math>BQ = \left(\frac{8-x}{2}\right)</math> and <math>CQ = \left(\frac{6-x}{2}\right)</math>. By the Pythagorean Theorem, <math>BQ^2+CQ^2=x^2</math>, or <math>\left(\frac{8-x}{2}\right)^2+\left(\frac{6-x}{2}\right)^2 = x^2</math>. Multiplying this out, we have <math>x^2 = \frac{64-16x+x^2+36-12x+x^2}{4}</math>. Simplifying, <math>-2x^2-28x+100=0</math>. Dividing both sides by <math>-2</math> gives <math>x^2+14x-50=0</math>. Therefore, using the quadratic formula, we have <math>x=-7 \pm 3\sqrt{11}</math>. Since lengths are always positive, then <math>x=-7+3\sqrt{11} \ | + | Let <math>AB</math>, or the side of the octagon, be <math>x</math>. Then, <math>BQ = \left(\frac{8-x}{2}\right)</math> and <math>CQ = \left(\frac{6-x}{2}\right)</math>. By the [[Pythagorean Theorem]], <math>BQ^2+CQ^2=x^2</math>, or <math>\left(\frac{8-x}{2}\right)^2+\left(\frac{6-x}{2}\right)^2 = x^2</math>. Multiplying this out, we have <math>x^2 = \frac{64-16x+x^2+36-12x+x^2}{4}</math>. Simplifying, <math>-2x^2-28x+100=0</math>. Dividing both sides by <math>-2</math> gives <math>x^2+14x-50=0</math>. Therefore, using the [[quadratic formula]], we have <math>x=-7 \pm 3\sqrt{11}</math>. Since lengths are always positive, then <math>x=-7+3\sqrt{11} \Rightarrow k+m+n=-7+3+11=\boxed{\textbf{(B)}\ 7}</math> |
~MrThinker | ~MrThinker |
Latest revision as of 19:37, 2 September 2024
Problem
In rectangle , and . Points and lie on , points and lie on , points and lie on , and points and lie on so that and the convex octagon is equilateral. The length of a side of this octagon can be expressed in the form , where , , and are integers and is not divisible by the square of any prime. What is ?
Solution 1
Let . Then .
Now notice that since we have .
Thus by the Pythagorean Theorem we have which becomes
Our answer is . (Mudkipswims42)
Solution 2
Denote the length of the equilateral octagon as . The length of can be expressed as . By the Pythagorean Theorem, we find that: Since , we can say that . We can discard the negative solution, so ~ blitzkrieg21
Solution 3
Let the octagon's side length be . Then and . By the Pythagorean theorem, , so . By expanding the left side and combining the like terms, we get . Solving this using the quadratic formula, , we use , , and , to get one positive solution, , so
Solution 4
Let , or the side of the octagon, be . Then, and . By the Pythagorean Theorem, , or . Multiplying this out, we have . Simplifying, . Dividing both sides by gives . Therefore, using the quadratic formula, we have . Since lengths are always positive, then
~MrThinker
Video Solution
~IceMatrix
See Also
2018 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
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